Given the periodic CT signal
$ \,x(t)=\frac{3\pi}{2}\cos(\frac{3\pi}{2}t+\pi)\sin(\frac{3\pi}{4}t+\frac{\pi}{2})\, $
compute its Fourier series coefficients.
Answer
We can rewrite the signal $ x(t) $ as
$ \,x(t)=\frac{3\pi}{2}(\frac{e^{j(\frac{3\pi}{2}t+\pi)}+e^{-j(\frac{3\pi}{2}t+\pi)}}{2})(\frac{e^{j(\frac{3\pi}{4}t+\frac{\pi}{2})}-e^{-j(\frac{3\pi}{4}t+\frac{\pi}{2})}}{2j})\, $
$ \,x(t)=\frac{3\pi}{8j}(e^{j\frac{3\pi}{2}t}e^{j\pi}+e^{-j\frac{3\pi}{2}t}e^{j\pi})(e^{j\frac{3\pi}{4}t}e^{j\frac{\pi}{2}}-e^{-j\frac{3\pi}{4}t}e^{j\frac{\pi}{2}})\, $
$ \,x(t)=\frac{3\pi}{8j}e^{j\pi}e^{j\frac{\pi}{2}}(e^{j\frac{3\pi}{2}t}+e^{-j\frac{3\pi}{2}t})(e^{j\frac{3\pi}{4}t}-e^{-j\frac{3\pi}{4}t})\, $
$ \,x(t)=-\frac{3\pi}{8}(e^{j(\frac{3\pi}{2}+\frac{3\pi}{4})t}-e^{j(\frac{3\pi}{2}-\frac{3\pi}{4})t}+e^{j(-\frac{3\pi}{2}+\frac{3\pi}{4})t}-e^{j(-\frac{3\pi}{2}-\frac{3\pi}{4})t})\, $
$ \,x(t)=-\frac{3\pi}{8}(e^{j\frac{9\pi}{4}t}-e^{j\frac{3\pi}{4}t}+e^{-j\frac{3\pi}{4}t}-e^{-j\frac{9\pi}{4}t})\, $
$ \,x(t)= -\frac{3\pi}{8}e^{j\frac{\pi}{4}t}+ \frac{3\pi}{8}e^{j\frac{3\pi}{4}t}- \frac{3\pi}{8}e^{-j\frac{3\pi}{4}t}+ \frac{3\pi}{8}e^{-j\frac{\pi}{4}t} \, $
